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Is the positive integer N a perfect square?

(1) The number of distinct factors of N is even.
(2) The sum of all distinct factors of N is even.

Originally posted by mbaMission on 02 Jun 2009, 04:18.
Last edited by Bunuel on 30 Jul 2012, 03:36, edited 1 time in total.
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1. The number of distinct factors of a perfect square is ALWAYS ODD. The reverse is also true: if a number has the odd number of distinct factors then it's a perfect square;

2. The sum of distinct factors of a perfect square is ALWAYS ODD. The reverse is NOT always true: a number may have the odd sum of its distinct factors and not be a perfect square. For example: 2, 8, 18 or 50;

3. A perfect square ALWAYS has an ODD number of Odd-factors, and EVEN number of Even-factors. The reverse is also true: if a number has an ODD number of Odd-factors, and EVEN number of Even-factors then it's a perfect square. For example: odd factors of 36 are 1, 3 and 9 (3 odd factor) and even factors are 2, 4, 6, 12, 18 and 36 (6 even factors);

4. Perfect square always has even powers of its prime factors. The reverse is also true: if a number has even powers of its prime factors then it's a perfect square. For example: $$36=2^2*3^2$$, powers of prime factors 2 and 3 are even.

NEXT:
There is a formula for Finding the Number of Factors of an Integer:

First make prime factorization of an integer $$n=a^p*b^q*c^r$$, where $$a$$, $$b$$, and $$c$$ are prime factors of $$n$$ and $$p$$, $$q$$, and $$r$$ are their powers.

The number of factors of $$n$$ will be expressed by the formula $$(p+1)(q+1)(r+1)$$. NOTE: this will include 1 and n itself.

Example: Finding the number of all factors of 450: $$450=2^1*3^2*5^2$$

Total number of factors of 450 including 1 and 450 itself is $$(1+1)*(2+1)*(2+1)=2*3*3=18$$ factors.

Back to the original question:

Is the positive integer N a perfect square?

(1) The number of distinct factors of N is even --> let's say $$n=a^p*b^q*c^r$$, given that the number of factors of $$n$$ is even --> $$(p+1)(q+1)(r+1)=even$$. But as we concluded if $$n$$ is a perfect square then powers of its primes $$p$$, $$q$$, and $$r$$ must be even, and in this case number of factors would be $$(p+1)(q+1)(r+1)=(even+1)(even+1)(even+1)=odd*odd*odd=odd\neq{even}$$. Hence $$n$$ cannot be a perfect square. Sufficient.

(2) The sum of all distinct factors of N is even --> if $$n$$ is a perfect square then (according to 3) sum of odd factors would be odd and sum of even factors would be even, so sum of all factors of perfect square would be $$odd+even=odd\neq{even}$$. Hence $$n$$ cannot be a perfect square. Sufficient.

Hope it helps.
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Interesting Question !!!

A few facts to review:

A perfect sqaure ALWAYS has an ODD number of factors, whose sum is ALWAYS ODD.

A perfect sqaure ALWAYS has an ODD number of Odd-factors, and EVEN number of Even-factors.

Using the above facts, you can conclude that both statements are sufficient to answer the question.
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mbaMission wrote:
Is the positive integer N a perfect square?

(1) The number of distinct factors of N is even.
(2) The sum of all distinct factors of N is even.

Note: A square never has even number of distinct factors. Also the sum of distinct factors of a square is never even.

(1) The number of distinct factors of N is even.

Suppose N = 4. It has 3 distinct factors: 1, 2 and 4.
Suppose N = 9. It has 3 distinct factors: 1, 3 and 9.
Suppose N = 16. It has 5 distinct factors: 1, 2, 4, 8, and 16.
Suppose N = 64. It has 7 distinct factors: 1, 2, 4, 8, 16, 32, and 64.
But that not the case. In fact, the case is opposite. So it is sufficient because N is not a square.

(2) The sum of all distinct factors of N is even.

If you follow the above pattern, you see 1 is always there. The sum of all distinct factors except 1 of N is even. If you add 1 on the even sum, that odd. So N is not a square.
But that not the case. In fact, the case is opposite. So it is sufficient because N is again not a square.

So D.

To validate the above premises, lets assume if N = 10. Its factors are 1, 2, 5 and 10 and their sum = 18. So 10 has 4 distinct factors and 18, which is even, sum.

If N = 15. Its factors are 1, 3, 5 and 15 and their sum = 24. So 15 has 4 distinct factors and even sum of its distinct factors.

If N = 30. Its factors are 1, 2, 3, 5, 6, 10, 15 and 30 and their sum = 40. So 15 has 8 distinct factors and even sum of its distinct factors. HTH.

Note: It would be better if you do not post OA or spoiler before any response from other members.
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Pkit wrote:

Is the positive integer N a perfect square?

(1) The number of distinct factors of N is even.
(2) The sum of all distinct factors of N is even.

I have got this question wrong , but I would argue with the OA provided by MGMAT.

OA D
.

1) The number of distinct factors of N is even.

Suppose N = 4. It has 3 distinct factors: 1, 2 and 4.
Suppose N = 9. It has 3 distinct factors: 1, 3 and 9.
Suppose N = 16. It has 5 distinct factors: 1, 2, 4, 8, and 16.
Suppose N = 64. It has 7 distinct factors: 1, 2, 4, 8, 16, 32, and 64.
But that not the case. In fact, the case is opposite. So it is sufficient because N is not a square.

(2) The sum of all distinct factors of N is even.

If you follow the above pattern, you see 1 is always there. The sum of all distinct factors except 1 of N is even. If you add 1 on the even sum, that odd. So N is not a square.
But that not the case. In fact, the case is opposite. So it is sufficient because N is again not a square.

So D.

PS: A perfect square always have odd number of factors, for e.g a integer $$n$$ and its square $$n^2$$

Now, $$n$$ will have always have even number of factors, (take any number and you will realise that factors come in pairs), now $$n^2$$ will have all factors which $$n$$ has + one more which is $$n^2$$
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If you don't know the tricks about perfect squares, it will take you a little time to demonstrate :

(1) SUFFICIENT: The factors of any number N can be sorted into pairs that multiply to give N. (For instance, the factors of 24 can be paired as follows: 1 and 24; 2 and 12; 3 and 8; 4 and 6.) However, if N is a perfect square, one of these ‘pairs’ will consist of just one number: the square root of N. (For example, if N were 49, it would have the factor pair 7 × 7) Since all of the other factors can be paired off, it follows that if N is a perfect square, then N has an odd number of factors. (If N is not a perfect square, then all of its factors can be paired off, so it will have an even number of factors.) This statement then implies that N is not a perfect square.

(2) SUFFICIENT: Let N be a perfect square. If N is odd, then all factors of N are odd. Therefore, by the above reasoning, N has an odd number of odd factors, so their sum must be odd. If N is even, let M be the product of all the odd prime factors (as many times as they appear in N – not distinct) of N, which is also a perfect square. (For instance, if N = 100, then M =5 × 5 = 25.) Then the sum of factors of M is odd, by the above reasoning. Furthermore, all other factors of N (i.e., that don’t also divide M) are even. The sum total is thus odd + even = odd.
Therefore, the sum of the factors of any perfect square is odd, so this statement implies that N is not a perfect square.
This statement can also be investigated by trying several cases of even perfect squares (4, 16, 36, 64, 100), noting that in each case the sums of the factors are odd, and generalizing.

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Fijisurf wrote:
Barkatis wrote:
(2) SUFFICIENT: Let N be a perfect square. If N is odd, then all factors of N are odd. Therefore, by the above reasoning, N has an odd number of odd factors, so their sum must be odd. If N is even, let M be the product of all the odd prime factors (as many times as they appear in N – not distinct) of N, which is also a perfect square. (For instance, if N = 100, then M =5 × 5 = 25.) Then the sum of factors of M is odd, by the above reasoning. Furthermore, all other factors of N (i.e., that don’t also divide M) are even. The sum total is thus odd + even = odd.
Therefore, the sum of the factors of any perfect square is odd, so this statement implies that N is not a perfect square.
This statement can also be investigated by trying several cases of even perfect squares (4, 16, 36, 64, 100), noting that in each case the sums of the factors are odd, and generalizing.

I am having trouble understanding this reasoning. If we are interested in sum of factors, why are you talking about product of odd prime factors?
Thanks.

Take an example N = $$2^4 * 3^2 *5^4$$
If we write down its factors, we get: 1, 2, 3, 4, 5, 6, 8, 10, 12..... total 75 factors.
Lets just consider the odd factors i.e. factors made by $$3^2 *5^4$$ (including 1)
These will be a total of 3*5 = 15 odd factors.
When you add an odd number of odd factors, result will be an odd number.
The rest of the factors of N will be even i.e. they will include at least one 2. their sum will definitely be even because no matter how many even numbers you add, you always get an even result.
Adding an odd number to an even number, you will get an odd number. Hence, sum of all factors of a perfect square will always be odd.
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VeritasPrepKarishma wrote:
Fijisurf wrote:
Barkatis wrote:
(2) SUFFICIENT: Let N be a perfect square. If N is odd, then all factors of N are odd. Therefore, by the above reasoning, N has an odd number of odd factors, so their sum must be odd. If N is even, let M be the product of all the odd prime factors (as many times as they appear in N – not distinct) of N, which is also a perfect square. (For instance, if N = 100, then M =5 × 5 = 25.) Then the sum of factors of M is odd, by the above reasoning. Furthermore, all other factors of N (i.e., that don’t also divide M) are even. The sum total is thus odd + even = odd.
Therefore, the sum of the factors of any perfect square is odd, so this statement implies that N is not a perfect square.
This statement can also be investigated by trying several cases of even perfect squares (4, 16, 36, 64, 100), noting that in each case the sums of the factors are odd, and generalizing.

I am having trouble understanding this reasoning. If we are interested in sum of factors, why are you talking about product of odd prime factors?
Thanks.

Take an example N = $$2^4 * 3^2 *5^4$$
If we write down its factors, we get: 1, 2, 3, 4, 5, 6, 8, 10, 12..... total 75 factors.
Lets just consider the odd factors i.e. factors made by $$3^2 *5^4$$ (including 1)
These will be a total of 3*5 = 15 odd factors.
When you add an odd number of odd factors, result will be an odd number.
The rest of the factors of N will be even i.e. they will include at least one 2. their sum will definitely be even because no matter how many even numbers you add, you always get an even result.
Adding an odd number to an even number, you will get an odd number. Hence, sum of all factors of a perfect square will always be odd.

I do understand now.
However still not sure why perfect squares have "EVEN number of Even-factors".
I understand that power of the only even prime factor "2" determines the number of even factors. So, if a number is a perfect square then all the powers will be even. Then by adding 1 to all powers we get odd numbers, which we multiply. Then the number of even factors will be odd.
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Fijisurf wrote:

I do understand now.
However still not sure why perfect squares have "EVEN number of Even-factors".
I understand that power of the only even prime factor "2" determines the number of even factors. So, if a number is a perfect square then all the powers will be even. Then by adding 1 to all powers we get odd numbers, which we multiply. Then the number of even factors will be odd.

OK. If you understand the theory above, will you agree that total number of factors of a perfect square will be odd (e.g. 75 above)? Also, all factors will be either even or odd. Will you also agree that total number of odd factors of perfect square will be odd? See the e.g. above $$3^2.5^4$$ will give 3*5 = 15 i.e. odd number of factors because powers are always even.
Then, the number of even factors should be Odd (75) - Odd(15) = Even(60).
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If you want the proper method:
We will obtain an even factor by multiplying all the 15 odd factors we obtained above by $$2 or 2^2 or 2^3 or 2^4$$.
Therefore, we will obtain 15*4 = 60 even factors. (We can take 2 in four ways)
If the number is a perfect square, power of 2 will always be even. (We do not use 4 + 1 here because to make it an even number, there has to be at least one 2.)
Therefore, even factors of a perfect square will always be even.
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You can see why this works if you build factor tables for a regular number and a perfect square:

30=
1*30=
2*15=
3*10=
5*6

Here all the factors come in pairs, so we will have an even number of factors.

36=
1*36=
2*18=
3*12=
4*9=
6*6
Here there is a pair of identical factors (that's the perfect square part), so we will always have an odd number.

Therefore, statement 1 tells us we don't have a perfect square. Our answer is "no"--sufficient.

As for the sum of the distinct factors, it's a bit more theoretical, but let's look at two possibilities:

1) The number is odd, in which case all of its factors are odd. If they come in pairs, each pair will add to make an even. (O+O=E) If the number is a perfect square, it will have an odd number of factors, leaving an extra odd number, which will make an odd sum. (E+O=O) Therefore, the sum of an odd perfect square's factors must be odd.
2) The number is even. If it is also a perfect square, it will have an even number of each prime factor, because the primes have to pair off (e.g. 100 = 10*10 = (2*5)(2*5)). All of its factors aside from 1 will be built from these primes, we will have an even number of evens (adding up to an even) and an even number of odds (also adding up to an even). Then 1 comes in and messes up all that harmony and makes the whole thing odd! Therefore, the sum of an even perfect square's factors must be odd.

So in either case, a perfect square's factors make an odd sum. Statement 2 gives an answer of "no," which is sufficient. The answer is D.

I hope this helps! You can definitely try this with some different numbers--it makes more sense when you play with a few examples.
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Bunuel wrote:
tingle15 wrote:
I have a doubt...

Consider N=18, Its factors are: 1, 2, 3, 6, 9, 18. The sum of factors is 39 which is odd... Am i missing something?

1. The number of distinct factors of a perfect square is ALWAYS ODD. The reverse is also true: if a number has the odd number of distinct factors then it's a perfect square;

I think we should mention that the unwritten assumption is that we are only talking about positive factors. If we include -ve factors, the number of factors of every integer is always even.
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glores1970 wrote:

I think we should mention that the unwritten assumption is that we are only talking about positive factors. If we include -ve factors, the number of factors of every integer is always even.

Factors are positive numbers only. On the exam if you have 'a has 4 factors', it means it has 4 positive factors.

The concepts of both the statements are discussed in detail in this post: http://www.veritasprep.com/blog/2010/12 ... t-squares/
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Bunuel wrote:
Pkit wrote:

Is the positive integer N a perfect square?

(1) The number of distinct factors of N is even.
(2) The sum of all distinct factors of N is even.

I have got this question wrong , but I would argue with the OA provided by MGMAT.

OA D
.

Probably the best way of solving would be making the chart of perfect squares and its factors to check both statements, but below is the algebraic approach if needed.

Couple of things:
1. Note that if $$n$$ is a perfect square powers of its prime factors must be even, for instance: $$36=2^2*3^2$$, powers of prime factors of 2 and 3 are even.

2. There is a formula for Finding the Number of Factors of an Integer:

First make prime factorization of an integer $$n=a^p*b^q*c^r$$, where $$a$$, $$b$$, and $$c$$ are prime factors of $$n$$ and $$p$$, $$q$$, and $$r$$ are their powers.

The number of factors of $$n$$ will be expressed by the formula $$(p+1)(q+1)(r+1)$$. NOTE: this will include 1 and n itself.

Example: Finding the number of all factors of 450: $$450=2^1*3^2*5^2$$

Total number of factors of 450 including 1 and 450 itself is $$(1+1)*(2+1)*(2+1)=2*3*3=18$$ factors.

3. A perfect square ALWAYS has an ODD number of Odd-factors, and EVEN number of Even-factors. For instance odd factors of 36 are 1, 3 and 9 (3 odd factor) and even factors are 2, 4, 6, 12, 18 and 36 (6 even factors).

Back to the original question:

Is the positive integer N a perfect square?

(1) The number of distinct factors of N is even --> let's say $$n=a^p*b^q*c^r$$, given that the number of factors of $$n$$ is even --> $$(p+1)(q+1)(r+1)=even$$. But as we concluded if n is a perfect square then powers of its primes $$p$$, $$q$$, and $$r$$ must be even, and in this case number of factors would be $$(p+1)(q+1)(r+1)=(even+1)(even+1)(even+1)=odd*odd*odd=odd\neq{even}$$. Hence $$n$$ can not be a perfect square. Sufficient.

(2) The sum of all distinct factors of N is even --> if n is a perfect square then (according to 3) sum of odd factors would be odd and sum of even factors would be even, so sum of all factors of perfect square would be $$odd+even=odd\neq{even}$$. Hence $$n$$ can not be a perfect square. Sufficient.

There are some tips about the perfect square:
• The number of distinct factors of a perfect square is ALWAYS ODD.
• The sum of distinct factors of a perfect square is ALWAYS ODD.
• A perfect square ALWAYS has an ODD number of Odd-factors, and EVEN number of Even-factors.
• Perfect square always has even number of powers of prime factors.

Hope it helps.

Thanks Bunuel for reminding us the very useful properties mentioned above.

For those interested, we can easily justify why the sum of the distinct divisors of a perfect square is odd.

If the number N is an odd perfect square, then all its divisors are odd. They come in pairs, $$(1,N), (d_1, N/d_1), (d_2,N/d_2)...$$, except $$\sqrt{N}$$ (we count it only once, its pair being itself). So, we have an odd number of odd divisors, whose sum will certainly be odd.

If N is an even perfect square, then N must be of the form $$N=2^{2n}M$$, where M is an odd perfect square. All the odd divisors of N are the divisors of M, and as we have seen above, their sum (and number) is odd. All the even divisors of N are obviously even, so again, the sum of all the divisors is odd.
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Re: Is the positive integer N a perfect square?  [#permalink]

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But in this particular question, it has not been specified if N is not equal to one.
Hence, N could be 1, perfect square, and have even number of factors namely 1 and -1.

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ShalabhAr wrote:
But in this particular question, it has not been specified if N is not equal to one.
Hence, N could be 1, perfect square, and have even number of factors namely 1 and -1.

Posted from my mobile device

Factor is a "positive divisor" (at least on the GMAT). For example the factors of 4 are 1, 2, and 4 ONLY.
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Re: Is the positive integer N a perfect square?  [#permalink]

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How is B sufficient? Sum of distinct factors of a perfect square is odd, but if n is 2, then the sum is also odd.
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ashish8 wrote:
How is B sufficient? Sum of distinct factors of a perfect square is odd, but if n is 2, then the sum is also odd.

The sum of distinct factors of a perfect square is ALWAYS ODD. (2) says that "the sum of all distinct factors of N is even", hence N is not a perfect square.
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ashish8 wrote:
How is B sufficient? Sum of distinct factors of a perfect square is odd, but if n is 2, then the sum is also odd.

(2) states: The sum of all distinct factors of N is even.
Since the sum of distinct factors of a perfect square must be odd, we can conclude that N is not a perfect square.
So, the answer to the question "Is N a perfect square?" is a definite NO.
Therefore, (2) sufficient.

Not only perfect squares have the sum of their distinct factors odd. As you mentioned, for 2, the sum of its factors is odd, and it is not a perfect square.
So, if a number is a perfect square, then the sum of its factors is necessarily odd, but the reciprocal is not true. Meaning, if the sum of the factors is odd, the number is not necessarily a perfect square, it might be or not. But if the sum of the distinct factors is even, then certainly the number cannot be a perfect square.
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